Question

an astronaut on a planet with no atmosphere drops a feather into a 6.21m deep crater and records that the feather falls freely for 1.69s. what is magnitude of free - fall acceleration on the planet? with what speed does the feather strike the bottom of the crater?

Explanation:

Step1: Use the free - fall displacement formula

The displacement formula for free - fall motion is $y = v_0t+\frac{1}{2}at^{2}$. Since the feather is dropped ($v_0 = 0$), the formula simplifies to $y=\frac{1}{2}at^{2}$, where $y$ is the displacement, $a$ is the acceleration, and $t$ is the time. We need to solve for $a$. Rearranging the formula gives $a=\frac{2y}{t^{2}}$.

Step2: Substitute the given values

We know that $y = 6.21m$ and $t = 1.69s$. Substituting these values into the formula $a=\frac{2y}{t^{2}}$, we have $a=\frac{2\times6.21}{(1.69)^{2}}$.
$a=\frac{12.42}{2.8561}\approx4.35m/s^{2}$.

Step3: Use the velocity - time formula for free - fall

The velocity - time formula for free - fall motion is $v = v_0+at$. Since $v_0 = 0$, the formula simplifies to $v = at$.

Step4: Substitute the values of $a$ and $t$

We know that $a\approx4.35m/s^{2}$ and $t = 1.69s$. Substituting these values into the formula $v = at$, we get $v=4.35\times1.69 = 7.35m/s$.

Answer:

Magnitude of free - fall acceleration: $4.35m/s^{2}$
Speed of the feather when it strikes the bottom: $7.35m/s$